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Is the ring of power series with coefficients in a field free as a module over the polynomials subring?

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    $\begingroup$ Obviously not (since $k[[x]]$ is $(x-1)$-divisible). $\endgroup$
    – YCor
    Commented May 9, 2022 at 21:04
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    $\begingroup$ Reading this title in the How Network questions list, I initially thought it was going to be a Tolkien question from SFF.stackexchange… $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented May 10, 2022 at 9:06

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No. Take some element $f$ of $K[x]$ which is not in the ideal $(x)$ and is not invertible. Then $$K[[x]] \otimes_{K[x]} K[x]/(f) \cong K[[x]]/(f) = 0$$ where the latter follows from the fact that since $f \notin (x)$, it becomes invertible in $K[[x]]$.

In particular, $K[[x]]$ is not faithfully flat over $K[x]$, so it, being a nonzero module, cannot be free over it.

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